The solution of the differential equation p²+q²=x+y \;{ is }\;

Correct Answer is : z= 23(a+x)32+ 23(y−a)32+bz=\;2/3(a+x)³/2}+\;2/3(y-a)³/2}+bz=32(a+x)23+32(y−a)23+b

z=\;2/3(a-x)³/2}+\;2/3(y+a)³/2}+b

z=\;2/3(a-x)³/2}+\;2/3(y-a)³/2}+b

z=\;2/3(a+x)³/2}+\;2/3(y+a)³/2}+b

z=\;2/3(a+x)³/2}+\;2/3(y-a)³/2}+b

Correct Answer is : z= 23(a+x)32+ 23(y−a)32+bz=\;2/3(a+x)³/2}+\;2/3(y-a)³/2}+bz=32(a+x)23+32(y−a)23+b

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UPSC ESE 2025 Electronics and Telecommunications Paper

Question : 144 of 150

Marks: +1, -0

p^{2}+q^{2}=x+y \;\text{ is }\;

Solution:

(d)

Given differential equation is

P^{2}+q^{2}=x+y

where

p=\;\frac{\partial z}{\partial x}, q=\;\frac{\partial z}{\partial y}

Let us take option (d) as solutions

z=\;\frac{2}{3}(a+x)^{\frac{3}{2}}+\;\frac{2}{3}(y-a)^{\frac{3}{2}}+b

p=\;\frac{\partial z}{\partial x}=\;\frac{2}{3}\cdot \frac{3}{2}(a+x)^{\frac{3}{2}-1}+0+0=\sqrt{a+x}

q=\;\frac{\partial z}{\partial y}=\;\frac{2}{3}\cdot \frac{3}{2}\cdot(y-a)^{\frac{3}{2}-1}=\sqrt{y-a}

Now

p^{2}+q^{2}=(a+x)+(y-a)=x+y

which is (1)

Hence, the correct option is (d).

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